Problem: Rewrite the function by completing the square. $f(x)=x^{2}-10x-96$ $f(x)=(x+$
Answer: We want to complete $x^2{-10}x$ into a perfect square. To do that, we should add $\left(\dfrac{{-10}}{2}\right)^2={25}$ to it: $x^2{-10}x+{25}=(x-5)^2$ In order to keep the expression equivalent, we add and subtract ${25}$, not forgetting the expression's constant term, $-96$ : $\begin{aligned} f(x)&=x^2-10x-96 \\\\ &=x^2-10x+{25}-96-{25} \\\\ &=(x-5)^2-96-25 \\\\ &=(x-5)^2-121 \end{aligned}$ In conclusion, after completing the square, the function is written as $f(x)=(x - 5)^2 - 121$ This is equivalent to $f(x)=(x+{-5})^2-121$